/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 Find the sum of the first 50 ter... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the sum of the first 50 terms of the arithmetic sequence: \(-10,-6,-2,2, \dots\)

Short Answer

Expert verified
The sum of the first 50 terms of the given arithmetic sequence is 3400.

Step by step solution

01

Find the First Term (a1)

The first term of the sequence is \(a_1 = -10\)
02

Calculate the Common Difference (d)

The common difference could be found as the difference between any two successive terms. Here, the second term is -6 and the first term is -10, so difference \(d = -6 - (-10) = 4\)
03

Find the 50th Term (a50)

The 50th term of an arithmetic sequence can be found using the formula \(a_n = a_1 + (n-1) * d\). Substituting the known values in, \(a_{50} = -10 + (50-1) * 4 = 146\)
04

Calculate Sum of the terms

The sum of the first 50 terms can be calculated using formula \(S_n = \frac{n}{2} * (a_1 + a_n)\). Substituting the known values in, \(S_{50} = \frac{50}{2} * (-10 + 146) = 50 * 68 = 3400\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Arithmetic Sequence Sum
Understanding the summation of an arithmetic sequence is key when dealing with linear patterns in numbers. An arithmetic sequence is a list of numbers where the difference between consecutive terms is constant, known as the common difference. In the exercise, we find a sequence starting with -10 and increasing by 4 each time.

To find the sum of the first 'n' terms of such a sequence, one effective formula is \(S_n = \frac{n}{2} * (a_1 + a_n)\). This formula is derived from the concept that the sum of an arithmetic series is the average of the first and last term, multiplied by the number of terms. In the case of the first 50 terms of the given sequence, this results in a sum of 3,400. It's like finding the average of the first and last number and then realizing the rest of the numbers fit symmetrically in between to give us our total sum.
Common Difference
A key characteristic of an arithmetic sequence is its common difference 'd', which is the consistent interval between any two consecutive terms. To find the common difference, simply subtract one term from the next. For instance, with the terms -10 and -6, the common difference is \(d = -6 - (-10) = 4\).

This difference remains the same throughout the sequence, and it is this steady step that allows an arithmetic sequence to be predictable and hence easy to work with. The common difference not only helps in identifying an arithmetic sequence but is also instrumental when finding other terms in the sequence or calculating the sum of a certain number of terms.
Sequence Term Calculation
Calculating individual terms in an arithmetic sequence is straightforward once the first term and the common difference are known. The general formula to find the nth term, \(a_n\), in an arithmetic sequence is \(a_n = a_1 + (n-1) * d\). This formula embodies the pattern by stating that to reach the nth term, one should start at the first term \(a_1\) and add the common difference 'd' a specific number of times, which is one less than 'n'.

In our case, to find the 50th term, we calculate \(a_{50} = -10 + (50-1) * 4\), which gives us 146. This demonstrates how an arithmetic sequence grows and how any term within the sequence can be pinpointed without listing out all terms before it.

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Most popular questions from this chapter

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The bar graphs show changes in educational attainment for Americans ages 25 and older from 1970 to 2007 . Exercises \(61-62\) involve developing arithmetic sequences that model the data. (GRAPH CANT COPY) In \(1970,55.2 \%\) of Americans ages 25 and older had completed four years of high school or more. On average, this percentage has increased by approximately 0.86 each year. a. Write a formula for the \(n\) th term of the arithmetic sequence that models the percentage of Americans ages 25 and older who had or will have completed four years of high school or more \(n\) years after 1969 . b. Use the model from part (a) to project the percentage of Americans ages 25 and older who will have completed four years of high school or more by 2019 .

Follow the outline below and use mathematical induction to prove the Binomial Theorem: $$\begin{aligned}(a+b)^{n} &-\left(\begin{array}{c}n \\\0\end{array}\right) a^{n}+\left(\begin{array}{c}n \\\1\end{array}\right) a^{n-1} b+\left(\begin{array}{c}n \\\2\end{array}\right) a^{n-2} b^{2} \\\&+\cdots+\left(\begin{array}{c}n \\\n-1\end{array}\right) a b^{n-1}+\left(\begin{array}{c}n \\\n\end{array}\right) b^{n}\end{aligned}$$ a. Verify the formula for \(n-1\) b. Replace \(n\) with \(k\) and write the statement that is assumed true. Replace \(n\) with \(k+1\) and write the statement that must be proved. c. Multiply both sides of the statement assumed to be true by \(a+b .\) Add exponents on the left. On the right, distribute \(a\) and \(b,\) respectively. d. Collect like terms on the right. At this point, you should have $$\begin{array}{l}(a+b)^{k+1}-\left(\begin{array}{c}k \\\0\end{array}\right)a^{k+1}+\left[\left(\begin{array}{c}k \\\0\end{array}\right)+\left(\begin{array}{c}k \\\1\end{array}\right)\right] a^{k} b \\\\+\left[\left(\begin{array}{c}k \\\1\end{array}\right)+\left(\begin{array}{c}k \\\2\end{array}\right)\right] a^{k-1} b^{2}+\left[\left(\begin{array}{c}k \\\2\end{array}\right)+\left(\begin{array}{c}k \\\3\end{array}\right)\right] a^{k-2} b^{3} \\\\+\cdots+\left[\left(\begin{array}{c}k \\\k-1\end{array}\right)+\left(\begin{array}{c}k \\\k\end{array}\right)\right] a b^{k}+\left(\begin{array}{c}k \\\k\end{array}\right) b^{k+1}\end{array}$$ e. Use the result of Exercise 84 to add the binomial sums in brackets. For example, because \(\left(\begin{array}{l}n \\\ r\end{array}\right)+\left(\begin{array}{c}n \\ r+1\end{array}\right)\) $$\begin{aligned}&-\left(\begin{array}{l}n+1 \\\r+1\end{array}\right), \text { then }\left(\begin{array}{l}k \\\0\end{array}\right)+\left(\begin{array}{l}k \\\1\end{array}\right)-\left(\begin{array}{c}k+1 \\\1\end{array}\right) \text { and }\\\&\left(\begin{array}{l}k \\\1\end{array}\right)+\left(\begin{array}{l}k \\\2\end{array}\right)-\left(\begin{array}{c}k+1 \\\2\end{array}\right)\end{aligned}$$ f. Because \(\left(\begin{array}{l}k \\\ 0\end{array}\right)-\left(\begin{array}{c}k+1 \\ 0\end{array}\right)(\text { why? })\) and \(\left(\begin{array}{l}k \\\ k\end{array}\right)-\left(\begin{array}{l}k+1 \\ k+1\end{array}\right)\) (why?), substitute these results and the results from part (e) into the equation in part (d). This should give the statement that we were required to prove in the second step of the mathematical induction process.

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